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Math Help - sin(x) + cos(x)= 1/2

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    sin(x) + cos(x)= 1/2

    Id like to know the solution to

    sin(x) + cos(x)= 1/2

    rather than the answer, i'd like to know how to work it out since i got a wrong answer.

    thanks,
    neilofbodom
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    Re: sin(x) + cos(x)= 1/2

    Quote Originally Posted by neilofbodom View Post
    Id like to know the solution to

    sin(x) + cos(x)= 1/2

    rather than the answer, i'd like to know how to work it out since i got a wrong answer.

    thanks,
    neilofbodom
    sin(x) + cos(x)= 1/2

    (sin(x) + cos(x))^2= (1/2)^2

    sin(2x)=(1/4)-1

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    Re: sin(x) + cos(x)= 1/2

    First thing you will need to do is graph the function, to see how many solutions you are expecting. You need to square both sides of the function to solve this equation, and squaring could bring in extraneous solutions.

    \displaystyle \begin{align*} \sin{x} + \cos{x} &= \frac{1}{2} \\ (\sin{x} + \cos{x})^2 &= \left(\frac{1}{2}\right)^2 \\ \sin^2{x} + \cos^2{x} + 2\sin{x}\cos{x} &= \frac{1}{4} \\ 1 + \sin{2x} &= \frac{1}{4} \\ \sin{2x} &= -\frac{3}{4} \\ 2x &= \left\{ \pi + \arcsin{\left(\frac{3}{4}\right)}, 2\pi - \arcsin{\left(\frac{3}{4}\right)}\right\}  + 2\pi n\textrm{ where }n \in \mathbf{Z} \\ x &= \left\{\frac{\pi}{2} + \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} , \pi - \frac{1}{2}\arcsin{\left(\frac{3}{4}\right)} \right\} + \pi n\end{align*}

    Now disregard any extraneous solutions.
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    Re: sin(x) + cos(x)= 1/2

    Quote Originally Posted by neilofbodom View Post
    Id like to know the solution to

    sin(x) + cos(x)= 1/2

    rather than the answer, i'd like to know how to work it out since i got a wrong answer.

    thanks,
    neilofbodom
    This has no solution in the 3rd quadrant
    as sin(x) and cos(x) are negative there.

    We can examine whether there is a solution in the 1st quadrant, where both are positive.

    f(x)=sinx+cosx\Rightarrow\ f'(x)=cosx-sinx

    The turning point occurs when sin(x) = cos(x)

    \Rightarrow\ x=\frac{\pi}{4}

    f''(x)=-sinx-cosx

    hence the first quadrant contains a maximum as f''(x)<0,\;\;\;x=\frac{\pi}{4}

    More significantly, the tangent slope is always decreasing in the first quadrant,
    hence the minimum values in the first quadrant occur either side of the maximum at x=0,\;\;\;x=\frac{\pi}{2}

    where f(x) =1 in both cases.

    Hence, there is no solution in the first quadrant either.

    There are certainly solutions in the 2nd and 4th quadrants, since sin(x) and cos(x)
    have opposite sign there.

    We can use the fact that cosx=sin\left(\frac{\pi}{2}-x\right)

    coupled with the identity

    sinA+sinB=2sin\frac{A+B}{2}cos\frac{A-B}{2}

    to obtain

    sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2sin\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)

    \Rightarrow\ \frac{2}{\sqrt{2}}cos\left(x-\frac{\pi}{4}\right)=\frac{1}{2}

    \Rightarrow\ cos\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{4}

    x=cos^{-1}\frac{\sqrt{2}}{4}+\frac{\pi}{4}

    and this gives the 2nd quadrant solution.

    Try the 4th quadrant solution.
    Hint: You'll need to reverse the signs of cos(x) and sin(x)
    Last edited by Archie Meade; June 20th 2011 at 04:16 PM. Reason: correction
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    Re: sin(x) + cos(x)= 1/2

    Hello, neilofbodom!

    Another approach . . .


    \text{Solve: }\;\sin x + \cos x \:=\: \frac{1}{2}

    \text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}

    \text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}

    \text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}
    \text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}

    \text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)

    \text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424

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    Re: sin(x) + cos(x)= 1/2

    Quote Originally Posted by Soroban View Post
    Hello, neilofbodom!

    Another approach . . .



    \text{Divide by }\sqrt{2}: \;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{2\sqrt{2}}

    \text{Note that: }\:\sin\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}},\;\;\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}

    \text{So we have: }\;\underbrace{\cos\tfrac{\pi}{4}\sin x + \sin\tfrac{\pi}{4}\cos x} \;=\;\frac{1}{2\sqrt{2}}
    \text{Then: }\qquad\qquad\quad\sin\left(x + \tfrac{\pi}{4}\right) \;=\;\frac{1}{2\sqrt{2}}

    \text{Hence: }\;x + \frac{\pi}{4} \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right)

    \text{Therefore: }\;x \;=\;\sin^{-1}\!\left(\frac{1}{2\sqrt{2}}\right) - \frac{\pi}{4} \;\approx\;-0.424

    Of course, to get all possible solutions...

    \displaystyle \begin{align*} \sin{\left(x + \frac{\pi}{4}\right)} &= \frac{\sqrt{2}}{4} \\ x + \frac{\pi}{4} &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)}, \pi - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n\textrm{ where }n\in\mathbf{Z} \\ x &= \left\{\arcsin{\left(\frac{\sqrt{2}}{4}\right)} - \frac{\pi}{4}, \frac{3\pi}{4} - \arcsin{\left(\frac{\sqrt{2}}{4}\right)}\right\} + 2\pi n \end{align*}
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    Re: sin(x) + cos(x)= 1/2

    Quote Originally Posted by neilofbodom View Post
    Id like to know the solution to

    sin(x) + cos(x)= 1/2

    rather than the answer, i'd like to know how to work it out since i got a wrong answer.

    thanks,
    neilofbodom
    You could also use simple geometry as shown in the attachment,
    using the fact that the sum of 2 sides of a triangle is greater than the 3rd side.
    This rules out the 1st and 3rd quadrants.

    Then you can find the 2nd quadrant solution using

    sinx+cosx=sinx+sin\left(\frac{\pi}{2}-x\right)=2cos\frac{\pi}{4}cos\left(x-\frac{\pi}{4}\right)

    to get

    x=arccos\left(\frac{\sqrt{2}}{4}\right)+\frac{\pi}  {4}

    Then to get the 4th quadrant solution, rotate the triangle to the position
    shown in the attachment to get

    x=\frac{9\pi}{4}-arccos\left(\frac{\sqrt{2}}{4}\right)
    Attached Files Attached Files
    Last edited by Archie Meade; June 21st 2011 at 01:22 PM. Reason: corrected attachment
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